Consider a system containing a negatively charge poin `(pi, m_(pi) = 273^(@)m_(e))` orbital around a staionary nucleus of atomic number Z .The total energy `(E_(n))` of ion is half of its potential energy `(PE_(n))` in nth sationary state .The motion of the poin can be assumed to be in a uniform circular notion with centripents force given by the force of attaraction between the positive uncless and the point .Assume that point revolves only in the stationary satte defined by the quantisation of its angular momentum about the nucless as Bohr's model
The wavelength `(lambda_(n))` of the pion orbital in nth stationarry state is ggiven by :

Options

A. `lambda_(pi) prop (n)/(m_(pi) z)`
B. `lambda_(pi) prop (m pi n)/(z)`
C. `lambda_(pi) prop (m pi z)/(n)`
D. `lambda_(pi) alpha (z)/(m_(pi) n)`

Explanation

To solve the problem, we need to derive the expression for the wavelength \( \lambda_n \) of the pion orbital in the nth stationary state based on the given conditions. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the relationship between wavelength and momentum
The wavelength \( \lambda \) of a particle can be expressed using de Broglie's relation:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

### Step 2: Express momentum in terms of mass and velocity
The momentum \( p \) of the pion can be expressed as:
\[ p = mv \]
where \( m \) is the mass of the pion and \( v \) is its velocity. Therefore, we can rewrite the wavelength as:
\[ \lambda = \frac{h}{mv} \]

### Step 3: Relate velocity to atomic number and quantum number
According to the Bohr model, the velocity \( v \) of an electron (or in this case, a pion) in a circular orbit is proportional to the atomic number \( Z \) and inversely proportional to the principal quantum number \( n \):
\[ v \propto \frac{Z}{n} \]
This can be expressed as:
\[ v = k \frac{Z}{n} \]
where \( k \) is a constant.

### Step 4: Substitute velocity into the wavelength equation
Substituting the expression for \( v \) into the wavelength equation, we get:
\[ \lambda = \frac{h}{m \left(k \frac{Z}{n}\right)} = \frac{h n}{k m Z} \]

### Step 5: Identify the proportionality
From the derived equation, we can see that the wavelength \( \lambda \) is proportional to \( n \) and inversely proportional to \( mZ \):
\[ \lambda \propto \frac{n}{mZ} \]

### Conclusion
Thus, the wavelength \( \lambda_n \) of the pion orbital in the nth stationary state is given by:
\[ \lambda_n \propto \frac{n}{mZ} \]

### Final Answer
The correct expression for the wavelength \( \lambda_n \) of the pion orbital in the nth stationary state is:
\[ \lambda_n \propto \frac{n}{mZ} \]

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